Question

If $f(x)=\begin{cases}\log \left(\sec ^{2} x\right)^{\cot ^{2} x}, & \text { for } x \neq0 \\ K, & x=0\end{cases}$
is continuous at $x =0$ then $K$ is

• A. $e^{-1} $
• B. $1$
• C. $e$
• D. $0$

Answer 1

Answer: (B)

Since, $f$ is continuous at $x=0$
$\therefore f(0)= \displaystyle\lim _{x \rightarrow 0} \log \left(\sec ^{2} x\right)^{\cot ^{2} x}$
Therefore we have,
$K=\displaystyle\lim _{x \rightarrow 0} \cot ^{2} x \cdot \log \left(\sec ^{2} x\right)$
$=\displaystyle\lim _{x \rightarrow 0} \cot ^{2} x \cdot \log \left(1+\tan ^{2} x\right)$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^{2} x\right)}{\tan ^{2} x} $
$\therefore K=1 \ldots \ldots . \displaystyle\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$